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I'm new in electrical stuff, but I know that a transistor has a voltage drop of around 0.7 V. So I have the following circuit:

schematic

(Hope it's correct): I want to trigger the transistor if the base get's flow from 3.3 V. My question: How do i calculate the voltage drop?

$$\frac{5~V - 0.7~V}{50~\Omega} = 0.086~A$$ $$\frac{3.3~V - 0.7~V}{500~\Omega} = 0.0052~A$$

So total at the ermitter:

$$0.086~A + 0.0052~A = 0.0912~A = 91.2~mA$$

Is this correct? Or it is only

$$\frac{5~V - 0.7~V}{50~\Omega} = 0.086~A$$ $$\frac{3.3~V}{500~\Omega} = 0.0066~A$$

or

$$\frac{5~V}{50~\Omega} = 0.1~A$$ $$\frac{3.3~V - 0.7~V}{500~\Omega} = 0.0052~A$$

I don't know where exactly the voltage drop should be in my calculation.

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    \$\begingroup\$ Your first line of calculation should probably use 0.2 V instead of 0.7 V, as you should be using a saturation voltage for \$V_{CE}\$ and not be using an Si diode PN junction voltage. To start. \$\endgroup\$ – jonk Jul 9 '17 at 16:30
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For the circuit you show, 0.7 only applies the base-emitter junction. The emitter-collector junction will attempt to affect current, rather than voltage.

So, the base current will be (3.3 - 0.7)/500, or 5.2 mA.

Calculating the collector current is harder. You must assume a value for the current gain, hfe or beta. Let's start by assuming an hfe of 100, which is a decent starting value for small-signal transistors.

5.2 mA times 100 is .52 amps. If this were true, the voltage across R1 would be .52 times 50, or 26 volts. And this obviously won't happen.

So it's likely that the transistor is not operating in the linear region, but is close to saturation. (The difference between linear and saturated operation is determined by whether the collector-emitter voltage is greater or less than the base-emitter voltage.) A good rule of thumb is that, in saturation, the gain of the transistor is less than 10 or 20, and the collector-emitter voltage will be in the range of 0.2 to 0.3 volts

Let's try 20. Then the current is .104 mA, and the voltage across R1 will be 5 volts. This is consistent with the circuit, and is a reasonable solution. The exact collector-emitter voltage cannot be determined easily, but in general the details won't matter. The current through R1 will be a bit less than 100 mA, and the voltage across the transistor will be about 0.2 to 0.3 volts.

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This is a grounded-emitter amplifier. That is, a common-emitter amplifier with RE=0. The gain of such circuit would be -RC/re, where re is the intrinsic emitter resistance, which is 25mV/Ic. For example, for a quiescent current of 1mA the gain is -Rc/re=-RC.Ic/25mV=-50*1/25=-2. But this is heavily dependent on the collector current. For your example, the gain will vary from -200 all the way down to zero. So in practice, it's better to avoid an emitter-grounded amplifier except when very small input signal swings are involved. People use emitter-follower with degeneration, where the voltage gain doesn't depend on the collector current.

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  • \$\begingroup\$ This is more of a saturated switching application of the transistor. Not an analog amplifier type of application. \$\endgroup\$ – mkeith Jul 10 '17 at 8:15
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The Collector-Emitter current (limit) is beta times Base-Emitter current. Beta is the gain of the transistor. Collector-Emitter voltage is (5V - 0.7V) / 50 Ohms = 86mA. Base-Emitter current is (3.3V - 0.7V) / 500 Ohms = 5.2mA. If beta were 200, then that Base-Emitter current could support driving up to about 1A of Collector-Emitter current.

You can generally assume a 0.2V drop across the Collector-Emitter junction. What you have to do is size your base resistor so that, for the beta of your transistor, at least enough current is flowing through the Base-Emitter junction in order to satisfy the demanded / desired / designed Collector-Emitter current.

This might be a good reference for you to take a look at.

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  • \$\begingroup\$ I use a BC547 NPN Transistor sparkfun.com/datasheets/Components/BC546.pdf could you show me how to calculate my example with this transistor? Sorry, but like i said, i'm new into this \$\endgroup\$ – M. Zim Jul 9 '17 at 16:49
  • \$\begingroup\$ What are you trying to accomplish exactly? \$\endgroup\$ – vicatcu Jul 9 '17 at 16:54
  • \$\begingroup\$ i'm learning how transistors worked and i builed this little setup. Now i'm calculating everything to see if i have understood everything. There is no "real goal". Just want to know about voltage drop. \$\endgroup\$ – M. Zim Jul 9 '17 at 16:58

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